Question

3x to the power 2 minus 4root 3 plus 4 equal to 0 . Solve using middle term splitting methods

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x=\frac{2}{\sqrt{3}}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {3x}^{2} - 4 \sqrt{3} x + 4 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies x = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {3x}^{2} - 4 \sqrt{3} x + 4 = 0 \\ \\ \tt: \implies {3x}^{2} - 2 \sqrt{3} x - 2\sqrt{3}x + 4 = 0 \\ \\ \tt: \implies \sqrt{3} x( \sqrt{3} x - 2) - 2( \sqrt{3} x - 2) = 0 \\ \\ \tt: \implies (\sqrt{3} x - 2)( \sqrt{3}x - 2) = 0 \\ \\ \tt: \implies (\sqrt{3} x - 2)^{2} = 0 \\ \\ \tt: \implies \sqrt{3} x - 2 = 0 \\ \\ \tt: \implies \sqrt{3} x = 2 \\ \\ \green{\tt: \implies x = \frac{2}{ \sqrt{3} } } \\ \\ \bold{Alternate \: method} \\ \tt: \implies {3x}^{2} - 4\sqrt{3} x + 4 = 0 \\ \\ \tt: \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \tt: \implies x = \frac{ - ( - 4 \sqrt{3} ) \pm \sqrt{( - 4 \sqrt{3} )^{2} - 4 \times 3 \times 4 } }{2 \times 3} \\ \\ \tt: \implies x = \frac{4 \sqrt{3} \pm \sqrt{16 \times 3 - 16 \times 3} }{6} \\ \\ \tt: \implies x = \frac{4 \sqrt{3} }{6} \\ \\ \green{\tt: \implies x = \frac{2}{ \sqrt{3} } }$

Answer 2

$\pink{ \huge \bold{Answer : }} \\ \green{\tt \therefore x = \frac{2}{ \sqrt{3} } }$

• From given question :

$\green{ \huge \bold{Step-by-step \: explanation}} \\ \\ \tt \to {3x}^{2} - 4 \sqrt{3} x + 4 = 0 \\ \\ \text{solving \: by \: middle \: term \: spliting} \\ \tt \to {3x}^{2} - 2\sqrt{3} x - 2 \sqrt{3} x+ 4 = 0 \\ \\ \tt \to \sqrt{3}x( \sqrt{3} x - 2) - 2( \sqrt{3} x - 4) = 0 \\ \\ \tt \to (\sqrt{3} x - 2)^{2} = 0 \\ \\ \tt \to \sqrt{3} x - 2 = 0 \\ \\ \tt \to \sqrt{3} x = 2 \\ \\ \green{\tt \to x = \frac{2}{ \sqrt{3} } } \\ \\ \green{\tt{ \huge\therefore x = \frac{2}{ \sqrt{3} } }}$